I've recently been introduced to calculus and ran into a rational function that seems to be somewhat

Akira Huang

Akira Huang

Answered question

2022-05-29

I've recently been introduced to calculus and ran into a rational function that seems to be somewhat complicated for my understanding, which is shown in the form:
f ( x ) = ( x a ) ( x b ) ( c a ) ( c b ) a + ( x b ) ( x c ) ( a b ) ( a c ) b + ( x a ) ( x c ) ( b a ) ( b c ) c
Bearing in mind that when f(a)=b, f(b)=c and f(c)=a due to substitution.
However, the part I don't understand is the simplification of the equation into the quadratic form as follows:
1 ( a b ) ( a c ) ( b c ) [ ( a b ) a + ( b c ) b + ( c a ) c ] x 2 + [ ( a b ) b 2 + ( b c ) c 2 + ( c a ) a 2 ] x + [ ( a b ) a 2 + ( b c ) b 2 c + ( c a ) a c 2 ]
Based on my algebraic knowledge, I know that the factorization formula is x 2 + ( a + b ) x + a b
My question is how did the coefficients and rearranged dominator of the original equation come about? Can anyone detail it for me, please?

Answer & Explanation

Lavizzariym

Lavizzariym

Beginner2022-05-30Added 10 answers

f ( x ) = ( x a ) ( x b ) ( c a ) ( c b ) a + ( x b ) ( x c ) ( a b ) ( a c ) b + ( x a ) ( x c ) ( b a ) ( b c ) c = ( x a ) ( x b ) ( a c ) ( b c ) a + ( x b ) ( x c ) ( a b ) ( a c ) b ( x a ) ( x c ) ( a b ) ( b c ) c = g ( x ) ( a b ) ( a c ) ( b c )
where
g ( x ) = ( x a ) ( x b ) a ( a b ) + ( x b ) ( x c ) b ( b c ) ( x a ) ( x c ) c ( a c )
The coefficient of x 2 of g(x) is just
The coefficient of x of g(x) is just
( a + b ) a ( a b ) ( b + c ) b ( b c ) + ( a + c ) c ( a c ) = a ( a 2 b 2 ) b ( b 2 c 2 ) + c ( a 2 c 2 ) = a 3 + a b 2 b 3 + b c 2 c 3 + c a 2 = b 2 ( a b ) + c 2 ( b c ) + a 2 ( c a )
Try to compute coefficient of 1 of g(x)

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